26 Worked Solutions

advertisement
25
Problem 9.2.A. A simple-span, uniformly –loaded
beam consists of a W 18  50 with Fy = 36 ksi. Find the
percentage gain in the limiting bending moment if a fully
plastic condition is assumed, instead of a condition
limited by elastic stress.
 S ´ X1 ´ 2 
Mn   x
 ´

Lb / ry


  ´
 245(1410) 2

 300 / 2.09


Elastic condition limit:
M = Fy  Sx = 36(88.9) = 3200 kip-in. = 267 kip-ft
489, 000
´
144
1


1
1
( X 1 )2 ´ X 2
2( Lb / ry ) 2
1410(0.0496)
2(300 / 2.09) 2
98, 600
41, 200
 3390´ 1.84
 6, 240 kip-in. or 520 kip-ft
Plastic condition limit:
M = Fy  Zx = 36(101) = 3640 kip-in. = 303 kip-ft
Gain in M = 303  267 = 36 kip-ft
% gain = 36/267  100 = 13.5%
Problem9.3.A. Determine the nominal bending moment
capacity ( Mn) for a W 30  90 made of A-36 steel and
the following unbraced lengths: (1) 5 ft, (2) 15 ft, (3) 30
ft.
(1) Lb = 5 ft. Table 9.1: Lp = 8.71 ft, Lr = 24.8 ft.
Problem 9.4.A. Design for flexure a simple beam 14 ft
[4.3 m] in length and having a total uniformly distributed
dead load of 13.2 kips [59 kN] and a total uniformly
distributed live load of 26.4 kips [108 kN].
 13.2 
wu  1.4(DL)  1.4 
  1.32 kips/ft
 14 
or
 13.2 
 26.4 
wu  1.2(DL)  1.6(LL)  1.2 
  1.6 

 14 
 14 
 1.14  3.02  4.16 kips/ft
Mn = Mp = Fy  Zx = 36(283) = 10,200 kip-in.
= 849 kip-ft
wu L2 4.16(14) 2

 102 kip-ft
8
8
M
102
Mn  u 
 113 kip-ft
Fb
0.9
(2) Lp  Lb  Lr, use:
Zx 
Mr = (Fy  Fr)  Sx = (36  10)(245) = 6370 kip-in.
= 531 kip-ft
From Table 9.1, try W 14  26
Since Lb  Lp, use:
 Lb - Lp
M n  M p - ( M p - M r )´ 
L - L
p
 r
 15 - 8.71 
 849 - (849 - 531) 

 24.8 - 8.71 
 725 kip-ft
(3) Lb  Lr, use:
26 Worked Solutions
Mu 



M n 113  12 in. 
3


  37.8 in.
Fy
36  1 ft 
New wu = 4.16 + 0.026 = 4.186
New Z x 
4.186
(37.8)  38.0
4.16
W 14  26 is still OK
Problem 9.4.C. A beam of 15-ft [4.6 m] length has three
concentrated loads of 6 kips, 7.5 kips, and 9 kips at 4 ft,
10 ft, and 12 ft [26.7 kN, 33.4 kN, and 40.0 kN at 1.2m,
3 m, and 3.6 m], respectively, from the left-hand support.
Design the beam for flexure.
Mn 
M u 78.5

 87.2 kip-ft
Fb
0.9
Zx 
M n 87.2(12)

 29.1 in.3
Fy
36
Pu = 1.2(dead load) + 1.6(live load)
From Table 9.1, try a W 12  22
Live loads: Pu1 = 1.6(6) = 9.6 kips, Pu2 = 1.6(7.5) = 12
kips, Pu3 = 1.6(9) = 14.4 kips.
Add weight of beam as a distributed load:
From analysis of the beam, Mu = 81.6 kip-ft
Mn 
Zx 
M u 81.6

 90.7 kip-ft
Fb
0.9
M n 90.7(12)

 30.2 in.3
Fy
36
From Table 9.1, select: W 10  26
wu
= 1.2(0.022) = 0.0264 kip/ft
M u  0.385 kip-ft (under point load)
New total Mn = 87.6 kip-ft
Zx 
Add weight of beam as a uniformly distributed load.
wu
= 1.2(dead load) = 1.2(0.026) = 0.0312 kips/ft
Mu 
wu L2 0.0312(15)2

 0.88 kip-ft
8
8
Mn 
M u 0.88

 0.98 kip-ft
Fb
0.9
New total Mn = 90.7 + 0.98 = 91.69 kip-ft
New required Zx = 30.5 in.3 , selection still OK.
Problem 9.4.E. Design for flexure a beam 12 ft [3.6 m]
in length, having a uniformly distributed dead load of 1
kip/ft [14.6 kN/m], a uniformly distributed live load of 1
kip/ft [14.6 kN/m], and a concentrated load of 8.4 kips
[37.4 kN] a distance of 5 ft [1.5 m] from one support.
M u 0.385

 0.427 kip-ft
Fb
0.9
Mn 
M n 87.6(12)

 29.3 in.3 . W 12 ´ 22 still OK
Fy
36
Problem 9.4.G. A steel beam 16-ft [4.9 m] long has a
uniformly distributed dead load of 100 lb/ft [1.46 kN/m]
and a uniformly distributed live load of 100 lb/ft [1.46
kN/m] extending to 10 ft [3 m] from the left support. In
addition, there is a concentrated live load of 8 kips [35.6
kN] at 10 ft [3 m] from the left support. Design the
beam for flexure.
wu = 1.2(DL) + 1.6(LL) = 1.2(0.1) + 1.6(0.1) = 0.28
kips/ft
Pu = 1.6(8) = 12.8 kips
From the beam analysis:
Mu = 54.6 kip-ft (under the point load)
Mn 
M u 54.6

 60.7 kip-ft
Fb
0.9
Zx 
M n 60.7(12)

 20.2 in.3
Fy
36
wu = 1.4(DL) = 1.4(1) = 1.4 kips/ft
or, wu = 1.2(DL) + 1.6(LL) = 1.2(1) + 1.6(1) = 2.8
kips/ft
From Table 9.1, try W 10  19
Pu = 1.4(DL) = 1.4(8.4) = 11.9 kips
Beam weight: wu = 1.2(0.019) = 0.0228 kips/ft
From the beam load analysis:
Adjusted Zx = 20.5 in.3, selection still OK
Mu = 78.5 kip-ft (under the point load)
27
Problem 9.4.I. A cantilever beam is 12 ft [3.6 m] long
and has a uniformly distributed dead load of 600 lbs/ft
[8.75 kN/m] and a uniformly distributed load of 1000
lbs/ft [14.6 kN/m]. Design the beam for flexure.
wu = 1.2(DL) + 1.6(LL) = 2.32 kips/ft
Mu 
wu L2 2.32(12) 2

 167 kip-ft
2
2
Mn 
M u 167

 186 kip-ft
Fb
0.9
Zx 
M n 186(12)

 61.9 in.3
Fy
36
From Table 9.1, try W 16  36
 Lb - Lp 
M n  M p - (M p - M r ) 
 L - L 
p 
 r
 6 - .71 
 462 - (462 - 286) 
  457 kip-ft
 17.2- 5.71 
Close but OK, use the W 24  62.
(b) Lb = 10 ft. From Table 9.1, try a W 24 68.
Additional weight makes the required Mn = 455 kip-ft.
 Lb - Lp 
- Mr 
 L - L 
p 
 r
 10- 7.50 
=480- (849- 303) 
  451 kip-ft
 22.8- 7.50 
Mn  M p -
M
p
For beam weight, wu = 1.2(0.036) = 0.0432 kips/ft
As this is less than the required value, the shape does not
work. Use a W 16  77 or a W 24  76.
New Mn = 190 kip-ft, Zx = 63.3 in.3, selection still OK
(c) Lb = 15 ft. Based on previous work, try a W 24  76.
Problem 9.5.A. A W shape is to be used for a uniformly
loaded simple beam carrying a total live load of 50 kips
[222 kN] and a total dead load of 22 kips [98 kN] on a 30
ft [9.2 m] span. Select the lightest weight shape for
unbraced lengths of (a) 6 ft [1.83 m], (b) 10 ft [3.05 m],
(c) 15 ft [4.57 m].
Wu = 1.2 (DL) + 1.6 (LL)
= 1.2(22) + 1.6(50) = 106.4 kips
Mu 
WL 106.4(30)

 399 kip-ft
8
8
Mn 
M u 399

 443 kip-ft
Fb
0.9
Zx 
M n 443(12)

 148 in.3
Fy
36
(a) Lb = 6 ft. From Table 9.1, a W 24  62 is lightest
(Lp) is 5.71 ft or less (Lb). Try this shape and check for
buckling. Adding the weight of the beam, new values
are:
Wu = 108.6 kips, Mn = 452 kip-ft
Check for: LpLbLr , 5.71617.2
 15- 8 
M n  600- (600- 381) 
  500 kip-ft
 23.4- 8 
Which indicates that the shape is adequate.
(Note: This process is really laborious. It makes a case
for using some aid to shortcut the process. One aid is the
use of a computer program. Another aid consists of the
series of graphs similar to Fig. 9.6 which are provided in
the AISC Manual (Ref. 5).
28 Worked Solutions
Problem 9.5.C. A W shape is to be used for a uniformly
loaded simple beam carrying a total live load of 50 kips
[222 kN] and a total dead load of 22 kips [98 kN] on a 30
ft [9.14 m] span. Select the lightest weigh shape for
unbraced lengths of (a) 6 ft; (b) 10 ft; (c) 15 ft.
Which indicates that the chosen shape is OK.
Wu  1.2( DL)1.6( LL)  1.2(22)  1.6(50)  106.4 kips
 15- 8 
M n  600- (600- 381) 
  501 kip-ft
 23.4- 8 
Mu 
WL 106.4(30)

 399 kip-ft
8
8
Mn 
M u 399

 443 kip-ft
Fb
0.9
Zx 
M n 443(12)

 148 in.3
Fy
36
(c) unbraced length = 15 ft
Try the W 24  76, requires Mn = 454 kip-ft.
Which indicates that the shape is still OK for this
unbraced length.
Problem 9.6.A, B, C. Compute the shear capacity (vVn)
for the following beams of A36 steel: A, W 24  84, C,
W 10  19.
Case 1:
h 418

 69.7
tw
36
Case 2:
h 523

 87.2
tw
36
From Table 9.1, lightest shape is a W 24  62 if
unbraced length is 5.71 ft or less.
(a) unbraced length Lb = 6 ft Try the W 24  62 for the
case of LpLbLr, or 5.71617.2. Then
Add weight of beam at 1.2(62  30) = 2.23 kips.
A: From Table A.3 = d = 24.1 in., tw = 0.470 in., tf =
0.770 in.
New required Mn = 453 Kip-ft
h = d  2(tf) = 24.1  2(0.770) = 22.56
 Lb - L p 
M n  M p - (M p - M r ) 
 L - L 
p 
 r
 6 - 5.71 
 462 - (462 - 286) 

 17.2 - 5.71 
 458 kip-ft
As this is greater than the required value, the shape
chosen is OK.
(b) unbraced length Lb = 10 ft Same case as (a).
Trials will show W 24  62, and 68 do not work.
Try W 24  76. New total Wu = 109.1 kips.
New required M n 
109.1
(443)  454 kip-ft
106.4
h 22.56

 48.0
tw
0.47
Case 1 condition
Aw = d  tw = 24.1  0.470 = 11.33 in.2
vVn = 0.9  0.6  36  11.33 = 220 kips
C: h = 10.24  2(0.395) = 9.45 in.
h 9.45

 37.8, Case 1
tw 0.25
Aw = 10.24  0.25 = 2.56 in.2
vVn = 0.9  0.6  36  2.56 = 49.8 kips
 10- 8 
M n  600- (600- 381) 
  572 kip-ft
 23.4- 8 
29
Problem 9.7.A-D. Find the maximum deflection in
inches for the following simple beams with uniformly
distributed load. Find the values using: (a) the equation
for the beam: and (b) the curves in Fig. 9.11.
A: W 10  33, span = 18 ft, service load = 1.67 klf.
5wL4
5(1.67 /12)(18´ 12) 4

 0.794 in.
(a) D 
384 EI
384(29, 000)(171)
(b)  = 0.9 in.
C: W 18  46, span = 24 ft, service load = 2.29 klf
5wL4
5(2.29 /12)(24´ 12) 4

 0.829 in.
(a) D 
384 EI
384(29, 000)(712)
(b)  = 0.8 in.
Problem 9.8.A-H. For each of the following conditions
find (a) the lightest permitted shape and (b) the
shallowest permitted shape of A36 steel.
Span (ft)
A
C
E
G
16
36
18
42
Superimposed Load (klf)
Live
Dead
3
1
0.333
1
3
0.5
0.625
0.238
A Factored load = 1.2(3  16) + 1.6(3  16) = 134 kips
(a) W 16  57, Table load = 142 kips
Beam weight adds 1.2(0.057  16) = 1.09 kips, which is
not a problem for the selection.
(b) W 10  88, Table load = 153 kips
Beam weight adds 1.2(0.088  16) = 1.69 kips, selection
OK.
C Factored load = 1.2(0.278  36) + 1.6(0.833  36)
= 60 kips
Beam weight adds 1.2(0.055  36) = 2.16 kips.
Not critical for the selection.
(b) W 18  86, Table load = 112 kips.
Beam weight not critical.
E Factored load = 1.2(0.625  18) + 1.6(0.333  18)
= 23 kips
(a) W 12  16, Table load = 24.1 kips.
Beam weight = 1.2(0.016  18) = 0.35 kips.
Not critical for the shape.
(b) W 10  19, Table load = 25.9 kips
Beam weight = 1.2(0.019  18) = 0.41 kips
Not critical for shape.
G Factored load = 1.2(0.238  42 + 1.6(1  42) =
=79 kips
(a) W 24  76, Table load = 103 kips
Beam weight not critical.
(b) W 21  83, Table load = 101 kips
Beam weight not critical.
Problem 9.10.A. Open web steel joists are to be used for
a roof with a live load of 25 psf and a dead load of 20 psf
(not including joist weight) on a span of 48 ft. Joists are
4 ft on center and deflection under live load is limited to
1/360 of the span. Select the lightest joist.
Live load = 25(4) = 100 lbs/ft
Dead load = 20(4) = 80 lbs/ft (without joists)
Factored load = 1.2(80) + 1.6(100) = 256 lbs/ft
Try 26K7 at 10.9 lbs/ft, Table load = 282 kips
Joist weight not critical.
Load for deflection from Table is 100 lbs/ft, exactly what
is required.
(a) W 21  50, Table load = 66 kips
30 Worked Solutions
Problem 9.10.C. Open web steel joists are to be used for
a floor with a live load of 50 psf and a dead load of 45
psf (without the joists) on a span of 36 ft. Joists are 2 ft
on center and deflection is limited to 1/360 under live
load and to 1/240 of the span under total load. Select (a)
the lightest possible joist, and (b) the shallowest possible
joist.
Live load = 50(2) = 100 lbs/ft
Problem 10.4.A. Using Table 10.2, select a column
section for an axial dead load of 60 kips and an axial live
load of 88 kips if the unbraced height about both the x
and y axes is 12 ft. A36 steel is to be used and K is
assumed as 1.0.
Dead load = 45(2) = 90 lbs/ft, without the joists.
Pu = 1.2(DL) + (1.6(LL) = 1.2(60) + 1.6(88) = 213 kips
Total service load = 190 lbs/ft
KL = 1.0(12) = 12 ft
Total factored load = 1.2(90) + 1.6(100) = 268 lbs/ft
From Table 10.2, possible choices are:
For total load deflection ned (240/360)(190) = 127 lbs/ft
(a) Try 24K4 at 8.4 lbs/ft, Table load = 340 lbs/ft for
total factored load, 150 lbs/ft for deflection due to total
service load.
(b) Try 22K6 at 8.8 lbs/ft, Table load = 381 lbs/ft for
total factored load, and 153 lbs/ft for deflection due to
total service load.
Added joist weights not critical.
Problem 10.3.A. Determine the maximum factored axial
load for a W 10  49 column with an unbraced height of
15 ft. Assume K = 1.0.
Table A.3: A = 15.6 in.2, ry = 2.48 in.
Shape:
Design Load:
W 14  53
W 12  45
W 10  33
W 8  31
355 kips
301 kips
222 kips
214 kips
Problem 10.4.C. Same data as Problem 10.4.A, except
the dead load is 142 kips and the live load is 213 kips.
The unbraced height about the x-axis is 20 ft and the
unbraced height about the y-axis is 10 ft.
Pu = 1.2(DL) + 1.6(LL) = 1.2(142) + 1.6(213) = 511 kips
( KL) y  1.0(10)  10 ft
( KL) x 1.0(20)

 11.4 ft
rx / ry
1.75
KL 1.0(15´ 12)

 72.6 Table 10.1: Fc = 27.2 ksi
ry
2.48
( KL)¢y 
Pu = c  Fc  A = 0.85  27.2  15.6 = 361 kips
X-axis controls, from Table 10.2 possible choices are:
Problem 10.3.C. Determine the maximum factored axial
load for the column in Problem 10.3.A, if the conditions
are as shown in Figure 10.5 with L1 = 15 ft and L2 = 8 ft.
Table A.3: A = 15.6 in.2, rx = 5.23 in., ry = 2.48 in.
KL 1.0(96)

 38.7 ,
ry
2.48
KL 1.0(180)

 34.4
rx
5.23
Shape:
Design Load:
W 14  68
W 12  79
W 10  68
511 kips
631 kips
520 kips
Problem 10.4.E-H. Select the minimum size standard
weight steel pipe for an axial dead load of 20 kips, a live
load of 30 kips, and the following unbraced heights: (e) 8
ft; (g) 18 ft.
Y axis governs, from Table 10.1: Fc = 33.2 ksi
Pu = cFcA = 0.85(33.2)(15.6) = 440 kips
Pu = 1.2(20) + 1.6(30) = 72 kips
(e) 4 in. pipe, table load = 77 kips
(g) 6 in. pipe, table load = 104 kips
31
with an effective unbraced height of 12 ft. Find the
maximum factored axial load.
Problem 10.4.I. A structural tubing column, designated
as HHS 4  4  3/8, of steel with Fy = 46 ksi, is used
From Table 10.5, factored load = 98 kips
Problem 10.4.K. Using Table 10.5, select the lightest
structural tubing column to carry an axial dead load of 30
kips and a live load of 34 kips if the effective unbraced
height is 10 ft.
column factored axial load = 200 kips, factored beam
reaction = 30 kips, unbraced column height is 14 ft.
From Table 10.2, m = 1.8 for the 14 ft height.
Pu¢  Pu  (m ´ M ux )  (200  30)  (1.8´ 30´ 6 /12)
Pu = 1.2(30) + 1.6(34) = 90.4 kips
 257 kips
From Table 10.5, possible choices are:
Section:
Area:
HHS 6  6  3/16
HHS 5  5  ¼
HHS 4  4  3/8
3.98 in.2
4.3 in.2
4.78 in.2
Table Load:
131 Kips
130 kips
119 kips
6 in. tube is lightest.
Problem 10.4.M. A double-angle compression member
8 ft long is composed of two A36 steel angles 4  3  3/8
in. with long legs back-to-back. Determine the
maximum factored axial load for the angles.
From Table 10.6: y-axis load: 104 kips, x-axis load: 112
kips. y-axis governs, load is 104 kips.
Problem 10.4.O. Using Table 10.6, select a doubleangle compression member for an axial compression
dead load of 25 kips and a live load of 25 kips if the
effective unbraced length is 10 ft.
Pu = 1.2(25) + 1.6(25) = 70 kips
Possible choice from Table 10.6:
5  3.5  5/16, for x-axis the table load limit is 70 kips.
Pu 230

 0.89 > 0.2, therefore correct formula was used
Pu¢ 257
From Table 10.2, select W 12  45
Problem 10.5.C. Same as Problem 10.5.A, except axial
load is 485 kips, beam reaction is 100 kips, unbraced
height is 18 ft.
Pu¢  Pu  m ´ M ux  (485  100)  (1.6´ 100 ´ 6 /12)
= 665 kips
Pu 485

 0.73 > 0.2, therefore correct formula used
Pu¢ 665
From Table 10.2, select W 12  96
Problem 10.5.E. a 14-in. W shape is to be used for a
column that sustains bending on both axes. Select a trial
section for an unbraced height of 16 ft and the following
factored values: total axial load = 80 kips, Mx = 85 kipft, My = 64 kip-ft., unbraced height is 16 ft.
Pu¢  Pu  (m´ M ux )  (2m´ M uy )
= (80 kips) + (1.7 ´ 85) + (2 ´ 1.7 ´ 64) = 443 kips
Pu
80

 0.18 < 0.2, thus incorrect formula was used
Pu¢ 443
Pu 9
 [(m ´ M ux )  (2m ´ M uy )]
2 8
80 9

 [(1.7 ´ 85)  (2´ 1.7 ´ 64)]  448 kips
2 8
Pu¢ 
Problem 10.5.A. It is desired to use a 12-in. W shape for
a column to support a beam as shown in Fig. 10.7.
Select a trial size for the column for the following data:
From Table 10.2, select W 14  82
32 Worked Solutions
Problem 11.2.A. A bolted connection of the general
form shown in Fig. 11.6 is to be used to transmit a
tension force of 75 kips dead load and 100 kips live load
by using 7/8 in. A325 bolts and plates of A36 steel with
threads included in the planes of shear. The outer plates
are to be 8 in. wide and the center plate is to be 12 in.
wide. Find the required thickness of the plates and the
number of bolts needed if the bolts are placed in two
rows. Sketch the final layout of the connection.
Pu = 1.2(DL) + 1.6(LL) = 1.2(75) + 1.6(100) = 250 kips
From Table 11.1, design strength of a single bolt
experiencing double shear with threads included is 43.3
kips.
n
Vu
250

 5.77 or 6
FvVr 43.3
From Table 11.2, minimum edge distance is 1.5 in. and
minimum spacing is 2.625 in.
For the plates, required cross-sectional area is
P
250
As  u 
 7.72 in.2
Ft Fy 0.9(36)
For the outer plates:
t
As
7.72

 0.482 in., use 1/2 in.
2(width) 2(8)
Thus the tension on the net area is not critical.
For bearing:
vRn = 1.5(1.5)(0,6875)(58) = 89.7 kips
Thus bearing is not critical.
Check for block shear on the middle plate.
Tension: net w = 5  1 = 4 in.
Ae = 4(11/16) = 2.75 in.2
Ft Pn  Ft ´ Fu ´ Ae  0.75(58)(2.75)  120 kips
Shear: net w = 2(1.5  0.5) = 2 in.
Ae = 2(11/16) = 1.375 in.2
Fv Rn  Fv ´ Fu ´ Ae  0.75(58)(1.375)  59.8 kips
Net width = 8 2(1.0) = 6 in.
Combined resistance = 120 + 59.8 = 180 kips
Net area: Ae = 6(0.5) = 3 in.2 per plate
Allowable tension on the net area:
Problem 12.2.A-F. Using data from Table 12.1, select
the lightest steel deck for the following:
Ft Pn  Ft ´ Fu ´ Ae  0.75(58)(2´ 3)  261 kips
Thus the tension on the net area is not critical.
For bearing resistance of a single bolt:
vRn = 1.5  Lc  t  Fu
= 1.5(1.5)(0.5)(58) = 65.25 kips
As this is greater than the bolt capacity, bearing is not
critical
For the middle plate:
t
As
7.72

 0.642 in., use 11/16 in.
width
12
Net width = 12 – 2(1.0) = 10 in.
Net area: Ae = 10(11/16) = 6.875 in.2
Allowable tension on the net area:
tPn = 0.75  Fu  As
= 0.75(58)(6.875) = 299 kips
Span Condition: Total Load (psf):
Choice:
A
simple span, 7 ft
45
WR20
C
two-span, 8.5 ft
45
WR18
E
three-span, 6 ft
50
IR22
Download
Related flashcards
Neonatology

55 Cards

Pediatrics

67 Cards

Neonatology

54 Cards

Pediatricians

20 Cards

Create flashcards